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Let $\Sigma$ be a finite set. Let $F_\Sigma$ be the free group over $\Sigma$. Let $G$ and $H$ be finite index subgroups of $F_\Sigma$. Consider the sets $GH$ and $HG$. Is it always true that $GH=HG$? If not, could you provide a counter-example?

The motivation for this question is automata theory. The subgroups G and H each represents a finite deterministic permutation automata. If the proposition above is true, it says something about the structure of the product automata.

flag	asked Feb 1 at 19:47 Yonatan 31●3

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Pick a finite group $\Gamma$ and two subgroups $A,B\subseteq\Gamma$ such that $AB\neq BA$. Pick any free group $F$ and any surjective morphim $f:F\to\Gamma$, and consider the subgroups $f^{-1}(A)$ and $f^{-1}(B)$. – Mariano Suárez-Alvarez Feb 1 at 19:56

anton · Accepted Answer · 2013-02-03 16:20:16Z UTC

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No, ths is not true. Let $E$ be a finite group and let $\phi:F_\Sigma\to E$ be a surjective group homomorphism. Let $A,B\subset E$ any subgroups and let $G=\phi^{-1}(A)$ and $H=\phi^{-1}(B)$. If $GH=HG$ holds, then by applying $\phi$ we get $AB=BA$. So if the claim was true, then for every pair of subgroups $A,B$ of any finite group we would have $AB=BA$. Now it's easy to find a counterexample.

edited Feb 3 at 16:20

answered Feb 1 at 20:02

anton
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The sentence «So for every pair of subgroups $A$, $B$ of any finite group we have $AB=BA$.» is plainly false. Maybe you could rephrase this somehow? – Mariano Suárez-Alvarez Feb 1 at 21:30

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@Mariano: I think that the intended logic of the proof is: if the OP's statement were true, then <<...>> would be true. But (as he and you agree) it is plainly false, it is easy to find a counterexample. – Lee Mosher Feb 2 at 1:18

@Lee, I know :-) – Mariano Suárez-Alvarez Feb 2 at 19:14

@Mariano, I did so. – anton Feb 3 at 16:21

Let G and H be finite index subgroups of a free group. Does GH=HG?

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